以任意数量的参数作为参数的函数 [英] Function with any number of arguments as parameter

问题描述

我有方法eval它需要函数和参数列表，目前我正在为每个可能的函数写作案例。

I have method eval which takes List of Function and arguments, currently I am writing case for every possible function. How can I write it more generic?

implicit class Expression(p: Product) {
  def eval = p.productIterator.toList match {
    case (f: ((Any) => Any)) :: p1 :: Nil => f(p1)
    case (f: ((Any, Any) => Any)) :: p1 :: p2 :: Nil => f(p1, p2)
    case (f: ((Any, Any, Any) => Any)) :: p1 :: p2 :: p3 :: Nil => f(p1, p2, p3)
  }
}
def print = (x: Any) => println(">> " + x)
def add = (x: Any, y: Any) => x.asInstanceOf[Int] + y.asInstanceOf[Int]
(print, "test").eval
(add, 2, 3).eval

推荐答案

你可以使用无形式来写这样的东西，它具有类型安全的优点， （比较在 Expression.eval 中使用任意）。

You could write something like this using shapeless, which has the benefit that it is type safe and checked at compile time (compared with using Any in your Expression.eval).

import shapeless._
import ops.hlist._
import syntax.std.function._
import ops.function._

def eval[P <: Product, G <: HList, F, L <: HList, R](
  p: P
)(implicit 
  gen: Generic.Aux[P, G], 
  ihc: IsHCons.Aux[G, F, L], 
  ftp: FnToProduct.Aux[F, L => R]
) = { 
  val l = gen.to(p)
  val (func, args) = (l.head, l.tail)
  func.toProduct(args)
}

您可以用作：

Which you could use as :

def add(a: Int, b: Int) = a + b
val stringLength: String => Int = _.length

eval((add _, 1, 2))             // Int = 3
eval((stringLength, "foobar"))  // Int = 6
eval((stringLength, 4))         // does not compile

您也可以将它与如果一个case类遵循相同的格式（首先是一个函数，然后是该函数的正确参数）：

You can also use it with a case class if it follows the same format (first a function then the right arguments for that function) :

case class Operation(f: (Int, Int) => Int, a: Int, b: Int)
eval(Operation(_ + _, 1, 2))    // Int = 3
eval(Operation(_ * _, 1, 2))    // Int = 2

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